静态和动态作用域 [英] statically and dynamically scopes
问题描述
我正在处理这个问题,我得到了答案:
静态:20,16
Dynamically:20,100
是否正确?
下面(用Pascal语言)。
语言的输出是静态范围的?如果语言是动态作用域,输出是什么?
程序main;
x:integer;
过程f1(z:integer)
begin
return z * x;
end
过程f2(z:integer)
int x;
begin
x = 2;
return f1(z)* x;
end
begin / *主程序* /
x = 5;
print f1(4);
print f2(4);
end
?使用静态范围的Perl:
my $ x = 5;
sub f1($){
my $ z = shift;
return $ z * $ x;
}
sub f2($){
my $ z = shift;
my $ x = 2;
return f1($ z)* $ x;
}
print f1(4),\\\
;
print f2(4),\\\
;
我得到 20
, 40
。 ( 20
为 4 * 5
, 40
c $ c>(4 * 5)* 2 。)
取代所有 my $ c $用
local
s获取动态范围,我得到 20
, 16
。 (
20
为 4 * 5
, 16
不幸的是,由于你只是发表了你的结论,没有解释,我可以''' t指出你在哪里错了。 。 。
I'm working on this problem and I got the answers :
Statically: 20, 16
Dynamically: 20, 100
is that correct?
Consider the program below (in a Pascal like language). What is the output of the language is statically scoped? What is the output if the language is dynamically scoped?
Program main;
x: integer;
procedure f1(z: integer)
begin
return z * x;
end
procedure f2(z: integer)
int x;
begin
x = 2;
return f1(z) * x;
end
begin /* main program */
x = 5;
print f1(4);
print f2(4);
end
Why not try out both versions? Using Perl with static scope:
my $x = 5;
sub f1($) {
my $z = shift;
return $z * $x;
}
sub f2($) {
my $z = shift;
my $x = 2;
return f1($z) * $x;
}
print f1(4), "\n";
print f2(4), "\n";
I get 20
, 40
. (20
being 4 * 5
, 40
being (4 * 5) * 2
.)
Replacing all the my
s with local
s to get dynamic scope, I get 20
, 16
. (20
being 4 * 5
, 16
being (4 * 2) * 2
.)
Unfortunately, since you only posted your conclusions, no explanation, I can't point out where you went wrong . . .
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